Embibe Experts Solutions for Chapter: Kinematics, Exercise 2: Level 2

The position of a particle as a function of time is given by $\overrightarrow{r}\left(t\right)=A\cos \omega t\hat{\mathrm{i}}+B\sin \omega t\hat{\mathrm{j}}$ where $A$ and $B$ are two real positive constants with $A>B$. The orbit of the particle looks like-

The initial velocity of a projectile is $\left(2\hat{\mathrm{i}}+4\hat{\mathrm{j}}\right) \mathrm{m} {\mathrm{s}}^{-1}$, where $\hat{\mathrm{i}}$ and $\hat{\mathrm{j}}$ are along horizontal and vertical directions, respectively. For $g=10 \mathrm{m} {\mathrm{s}}^{-2}$, the equation of the trajectory will be,

$\int \left(\frac{\sqrt{x}}{2}+\frac{2}{\sqrt{x}}\right)dx$

$\int \left(8y-\frac{2}{y^{1/4}}\right)dy$

$\int 2x\left(1-x^{-3}\right)dx$

$\int \left(-5\sin t\right)dt$

If $y=x^2,$ then area of curve $y \text{vs} x$ from $x=0$ to $2$ will be 

In the cube of side $\text{'}a\text{'}$ shown in the figure, the vector from the central point of the face $ABOD$ to the central point of the face $BEFO$ will be :